Evaluate the definite integral. $\int^4_{3}\left(4+\dfrac{24}{x^2}\right)\,dx = $
Solution: First, use the power rule: $\begin{aligned}\int^4_{3}\left(4+\dfrac{24}{x^2}\right)\,dx~&= \int^4_{3}\left(4+24x^{-2}\right)\,dx \\&= ~(4x-24x^{-1})\Bigg|^4_{3}\end{aligned}$ Second, plug in the limits of integration: $[4\cdot4-24\cdot4^{-1}]-[4\cdot3-24\cdot3^{-1}] = 10-4 = 6$. The answer: $\int^4_{3}\left(4+\dfrac{24}{x^2}\right)\,dx = 6$